verifying log rules via elementary properties of an integral my attempt at this question so far,  
$$\int_1^x \frac{1}{t}dt +\int_1^y \frac{1}{t}dt= 2\int_1^x \frac{1}{t}dt+  \int_x^y \frac{1}{t}dt$$
But I am not sure hwo to prove it from elementary properties, I have also tried looking at the graph but the method doesn't seem clear to me.

 A: A basic property of integrals is the addition of integrals over adjacent intervals of integration:
$$\int_1^{xy}\frac{1}{t}dt=\int_y^{xy}\frac{1}{t}dt+\int_1^{y}\frac{1}{t}dt.$$
Substituting $t=yu$ into the first integral on the RHS, and to obtain the desired identity:
$$\int_y^{xy}\frac{1}{t}dt=\int_{y/y}^{xy/y}\frac{1}{yu}ydu=\int_1^x\frac{1}{u}du.$$
A: Assume $1<x$, $1<y$:
$$\int_1^{xy} \frac{1}{t}dt = \int_1^x \frac{1}{t}dt+  \int_x^{xy} \frac{1}{t}dt$$
Then in:
 $$\int_x^{xy} \frac{1}{t}dt$$ 
Use $u=\frac{t}{x}$: 
$$\int_x^{xy} \frac{1}{t}dt=\int_1^{y} \frac{1}{u}du$$ 
A: Hint: Substitute, $u=t/x$ in the second term. It becomes,
$$\int_x^{xy} \frac{{\rm d}u}{u}$$
A: $$∫_1^x\frac{1}{t} dt=\ln x- \ln 1=\ln x-0=\ln x$$
Also ,
$$∫_1^y\frac{1}{t} dt=\ln y-ln 1=\ln  y-0=\ln y$$
So 
$$∫_1^x\frac{1}{t} dt+∫_1^y\frac{1}{t} dt=\ln x+\ln y$$
We can provide the answer for $\ln x+\ln y$ we use the definition of $\ln x $,$\ln x=\frac{\log x}{\log e}
\ln x+\ln y=\frac{\log x}{\log e}+\frac{\log y}{\log e}=\frac{\log x+\log y}{\log e} =\frac{\log x+\log y}{1}=\ln xy$
$$\ln(xy)=∫_1^xy\frac{1}{t}dt$$
